Re: calculation of kmag
"Steven G. Johnson" <[email protected]>
| Newsgroups | gmane.comp.science.photonic-bands |
|---|---|
| Message-ID | <[email protected]> |
On Dec 15, 2009, at 3:52 PM, Oliver Willekens wrote:
> I wanted to plot the lightcone by plotting kmag versus k1 and k2 in a
> triangular lattice. However, it does not look like a cone(segment).
> Plotting it versus sqrt(k1^2+k2^2) works of course.
kmag is not sqrt(k1^2 + k2^2), except in a square lattice. The reason
is that k1 and k2 are in the reciprocal basis which is non-Cartesian
in a non-square lattice, and hence the Pythagorean theorem does not
apply.
In MPB, kmag is (vector3-norm (reciprocal->cartesian k))
> k1 k2 k3 kmag/2pi (comment)
> 0.3333 0.3333 0 0.3849 (K-point)
> 0.5000 0 0 0.5774 (M-point)
>
> However, in this basis, the magnitude of the vector Gamma-K should be
> 1/sqrt(3), no?
No. The correct answer is 2/3 / sqrt(3) = 0.384900179459751.
The reciprocal lattice vectors for your lattice are 2pi/a (1/sqrt(3),
+/- 1). Hence the K point, in Cartesian coordinates, is the sum of
these two vectors times 1/3, or 2pi/a (2/sqrt(3), 0) / 3. Hence the
length is 2/3 / sqrt(3) in units of 2pi/a.
See appendix B of our book if you need more explanation about
reciprocal lattices (http://ab-initio.mit.edu/book).
In general, when you plot kmag over the irreducible Brillouin zone
boundaries, the only portions that will be straight lines ("look like
a cone") will be the portions that correspond to straight lines from
the origin (Gamma).
Steven